Subject: Chemistry — Atomic Structure
Question: For $He^{+}$, a transition takes place from the orbit
of radius 105.8 pm to the orbit of radius 26.45 pm. The wavelength (in
nm) of the emitted photon during the transition is \_\_\_\_\_\_\_\_.
Bohr radius, a = 52.9 pm
Rydberg constant $R_{H} = 2.2 \times 10^{- 18}J$
Planck's constant, $h = 6.6 \times 10^{- 34}Js$
Speed of light, $c = 3 \times 10^{8}\ ms^{- 1}$
Answer: 30
Solution: $\frac{1}{\lambda} = \frac{R_{H}Z^{2}}{h.c}\left\lbrack \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right\rbrack$
$$\frac{1}{\lambda} = \frac{2.2 \times 10^{- 18} \times Z^{2}}{6.6 \times 10^{- 34} \times 3 \times 10^{8}}\left\lbrack \frac{1}{1^{2}} - \frac{1}{2^{2}} \right\rbrack$$
⇒
$\frac{2.2 \times 10^{- 18} \times 4}{6.6 \times 3 \times 10^{- 26}} \times \frac{3}{4}$
$$\lambda = \frac{6.6 \times 3 \times 10^{- 26}}{2.2 \times 3 \times 10^{- 18}}$$
⇒ $3 \times 10^{- 8} \Rightarrow 30 \times 10^{- 9}$
$$= 30\ nm$$